Question

Find the derivative of the following functions from first principle:
(i) −x (ii) (-x)-1 (iii) sin (x + 1) (iv) cos (x – \(\frac{\pi}{8}\))

Answer 1

(i) Let f(x) = -x. Accordingly, f(x+h)=-(x+h)
By first principle,
f'(x)=\(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
=\(\lim_{h\rightarrow 0}\) \(\frac{-(x+h)-(-x)}{h}\)
=\(\lim_{h\rightarrow 0}\) \(\frac{-x-h+x}{h}\)
=\(\lim_{h\rightarrow 0}\) \(\frac{-h}{h}\)
=\(\lim_{h\rightarrow 0}\)(-1)=-1

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(ii) Let f(x)=(-x)^-1 = \(\frac{1}{-x}\) = \(-\frac{1}{x}\) Accordingly, f(x+h)=\(-\frac{1}{x+h}\)
By first principle,
f'(x)=\(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[\(-\frac{1}{x+h}-(\frac{-1}{x})\)]
=\(\lim_{h\rightarrow 0}\)\(\frac{1}{h}\)[\(\frac{-1}{x+h}+\frac{1}{x}\)]
=\(\lim_{h\rightarrow 0}\)\(\frac{1}{h}\)[\(\frac{-x+(x+h)}{x(x+h)}\)]
=\(\lim_{h\rightarrow 0}\)\(\frac{1}{h}\)[\(\frac{-x+x+h}{x(x+h)}\)]
=\(\lim_{h\rightarrow 0}\)\(\frac{1}{h}\)[\(\frac{h}{x}\)(x+h)]
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{x}\)(x+h)
= \(\frac{1}{x}\).x - \(\frac{1}{x^2}\)

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(iii) Let f(x) = sin (x + 1). Accordingly, f(x+h)=sin(x+h +1)
By first principle,
f'(x)=\(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[sin(x+h+1) -sin(x+1)]
=\(\lim_{h\rightarrow 0}\) 1\(\frac{1}{h}\)[2cos(x+h+1+x+\(\frac{1}{2}\)) sin(x+h+1-x-1)]
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[ 2cos (2x + h+\(\frac{2}{2}\)) sin(\(\frac{h}{2}\))]
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[ 2cos (\(\frac{2x+h+2}{2}\)) sin(\(\frac{h}{2}\))/(\(\frac{h}{2}\))]
=cos(\(\frac{2x+0+2}{2}\)).1 [\(\lim_{h\rightarrow 0}\) \(\frac{sin\,x}{x}\) =1]
=cos(x+1)

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(iv) letf(x) cos(x-\(\frac{\pi}{8}\)). Accordingly, f(x+h)=cos(x+h -\(\frac{\pi}{8}\))
By first principle,
f'(x)=\(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
=\(\lim_{h\rightarrow 0}\)\(\frac{1}{h}\)[cos(x+h-\(\frac{\pi}{8}\))- cos(x-\(\frac{\pi}{8}\))]
=\(\lim_{h\rightarrow 0}\)\(\frac{1}{h}\)[\(\frac{-2sin(x+h-\frac{\pi}{8}+-\frac{\pi}{8})}{2\,sin(\frac{x+h-\frac{\pi}{8}-x+\frac{\pi}{8}}{2})}\)]
=\(\lim_{h\rightarrow 0}\)\(\frac{1}{h}\)[\(\frac{-2sin(2x+h-\frac{\pi}{4})}{2}\) \(sin \frac{h}{2}\)]
=\(\lim_{h\rightarrow 0}\)[-sin(\(\frac{2x+h-\frac{\pi}{4}}{2}\)) \(\frac{sin\frac{h}{2}}{\frac{h}{2}}\)]
=-sin(\(\frac{2x+0-\frac{\pi}{4}}{2}\)) .1
=-sin(x-\(\frac{\pi}{8}\))